Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x-9y &= -4 \\ 4x+3y &= -8\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $4x = -3y-8$ Divide both sides by $4$ to isolate $x$ $x = {-\dfrac{3}{4}y - 2}$ Substitute this expression for $x$ in the first equation. $8({-\dfrac{3}{4}y - 2}) - 9y = -4$ $-6y - 16 - 9y = -4$ Simplify by combining terms, then solve for $y$ $-15y - 16 = -4$ $-15y = 12$ $y = -\dfrac{4}{5}$ Substitute $-\dfrac{4}{5}$ for $y$ in the top equation. $8x-9( -\dfrac{4}{5}) = -4$ $8x+\dfrac{36}{5} = -4$ $8x = -\dfrac{56}{5}$ $x = -\dfrac{7}{5}$ The solution is $\enspace x = -\dfrac{7}{5}, \enspace y = -\dfrac{4}{5}$.